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3 years ago

Locate the complete verbal phrase and identify its type.

2 answers:

7 0

won
adjective

Verb phrases are verbs that may function as a predicate, adjective, or adverb. 

(a) "That he said" is an adjective modifying "word". However, this contains the s ubject"he" and the verb "said". It is a clause and NOT a phrase. Phrases can only have either a verb or a noun.
(b) There's only one verb "was" but it does not come with a complement, object, modifier, or other verb. Hence, it's NOT a verb phrase. 
(c) "Shall be" consists of the modal shall and the be-verb be. This is a perfect example of a verb phrase that functions as a VERB PHRASE. 
(d) "Roared" and "charged" are two verbs referring to different subjects. They do not come with a complement, object, modifier, or another verb. Hence, they're NOT a verb phrase. "As the bull charged" is a clause and not a phrase.

rodikova [14]3 years ago

5 0

The correct answer is:

Verbal Phrase: named Robert

Type: participial

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(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

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From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

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